The Existence of Ground State Solutions for Schrödinger-Kirchhoff Equations Involving the Potential without a Positive Lower Bound

In this paper, we study the following Schrödinger-Kirchhoff equation where V( x) ≥ 0 and vanishes on an open set of R ² and f has critical exponential growth. By using a version of Trudinger-Moser inequality and variational methods, we obtain the existence of ground state solutions for this problem.

• Schrödinger-Kirchhoff Equations
• Critical Exponential Growth
• Ground State Solution
• Degenerate Potential

1. Introduction

In this paper, we study the existence of ground state solutions for the following Schrödinger-Kirchhoff equation:

− ( a + b ∫ ℝ 2 | ∇ u | 2 d x ) Δ u + V ( x ) u = f ( u ) , u ∈ H 1 ( ℝ 2 ) , (1.1)

where a , b > 0 and the potential V ( x ) ≥ 0 satisfying:

(V₁) V ( x ) = 0 at B δ ( 0 ) and V ( x ) ≥ C 0 in ℝ 2 \ B 2 δ ( 0 ) for some C 0 , δ > 0 .

(V₂) There holds true;

sup x ∈ ℝ 2 V ( x ) = lim | x | → ∞ V ( x ) = γ > 0 , (1.2)

and the nonlinear term f ( t ) is a continuous function on ℝ . Moreover, we impose the following conditions on the nonlinearity f ( t ) ;

(f₀) f ( t ) = 0 for all t ≤ 0 ;

(f₁) critical exponential growth; there exists α 0 > 0 such that

lim | t | → + ∞ | f ( t ) | e α t 2 = { 0         for   a > a 0 , + ∞   for   a < a 0 ; (1.3)

(f₂) there exists μ > 4 such that

0 < μ F ( t ) = μ ∫ 0 t   f ( s ) d s ≤ t f ( t )   ∀ t ∈ ℝ \ { 0 } . (1.4)

(f₃) there exist t 0 > 0 and M 0 > 0 such that F ( t ) ≤ M 0 | f ( t ) | for any | t | ≥ t 0 ;

(f₄) f ( t ) = o ( t ) and f ( 0 ) = 0 ;

(f₅) f ( t ) ∈ C 1 ( ℝ ) and f ( t ) t 3 is increasing.

Without losing generality, we suppose that a = b = 1 . So we may rewrite problem (1.1) in the following form:

− ( 1 + ∫ ℝ 3 | ∇ u | 2 d x ) Δ u + V ( x ) u = f ( u ) , u ∈ H 1 ( ℝ 3 ) . (1.5)

Remark 1.1 The condition (f₂) implies that F ( t ) = o ( t 2 ) as t → 0 . Indeed, the condition (f₂) implies that

( F ( t ) t μ ) ′ > 0 , (1.6)

from which we can promptly obtain F ( t ) = o ( t 2 ) as t → 0 . From the condition (f₁), (f₂) and (f₄), we can get the following growth condition for f ( t ) ; for any ε > 0 and β 0 > α 0 , there exists C ε such that

| f ( t ) | ≤ ε | t | + C ε t μ ( e β 0 t 2 − 1 ) ,   ∀ t ∈ ℝ . (1.7)

From the condition (f₅), we can also easily check that the function f ( t ) t − 4 F ( t ) is increasing.

The corresponding Dirichlet problem for (1.1) on a smooth domain Ω ⊂ ℝ 2 ,

( − ( a + b ∫ Ω | ∇ u | 2 d x ) Δ u = f ( x , u ) x ∈ Ω , u = 0 x ∈ ∂ Ω , (1.8)

is related to the stationary analogue of the Kirchhoff equation

u t t − ( a + b ∫ Ω | ∇ u | 2 d x ) Δ u = f ( x , u ) , (1.9)

which was first proposed by Kirchhoff [1] as an extension of the classical D’Alembert’s wave equation for free vibrations of elastic strings. Problem (1.9) has attracted considerable attention after Lions [2] introduced an abstract framework to the problem.

The above problem is nonlocal as the appearance of the term ∫ Ω | ∇ u | 2 d x implies that (1.5) is not a pointwise identity. This phenomenon provokes some mathematical difficulties, which make the study of such class of problems particularly interesting. For more details on the physical and mathematical background of this problem to [2] [3] [4] [5].

Since we will work with critical exponential growth, we need to review the Trudinger-Moser inequality. For one thing, let Ω denote a smooth bounded domain in ℝ N ( N ≥ 2 ) . N. Trudinger [6] proved that there exists α > 0 such that W 0 1, N ( Ω ) is embedded in the Orlicz space L φ α ( Ω ) determined by the

Young function φ α = e α | t | N N − 1 . It was sharpened by J. Moser [7] who found the best exponent α n = n ω n − 1 1 n − 1 , where ω n − 1 is the surface measure of the unit

sphere in ℝ N . For another, the Trudinger-Moser inequality was extended for unbounded domains by D. M. Cao [8] in ℝ 2 and for any dimension N ≤ 2 by J. M. do Ó [9]. Moreover, J. M. do Ó et al. [10] established a sharp Concentration-compactness principle associated with the singular Trudinger-Moser inequality in ℝ N .

Many significant research results about (1.1) have been obtained. For example, in [11], X. Wu studied the nontrivial solutions and high energy solutions of problem (1.1) if V ( x ) has a positive constant lower bound and the nonlinearities term with 4-superlinear growth at infinity. In [12], the authors studied the following Schrödinger-Kirchhoff type equation

M ( ∫ ℝ 2 | ∇ u | 2 + V ( x ) u 2 d x ) ( − Δ u + V ( x ) u ) = A ( x ) f ( u )     in   ℝ 2 (1.10)

where M is a Kirchoff-type function and V ( x ) ≥ V 0 is a continuous function, A is locally bounded and the function f has critical exponential growth. Applying variational methods beside a new Trudinger-Moser type inequality, they get the of ground state solution. Moreover, in the the local case M ≡ 1 , they also get some relevant results. We emphasize that in these papers, the potential V ( x ) have a positive constant lower bound. Some studies of the Kirchhoff equation with critical exponential growth may refer [13] [14] [15] [16].

In [17], the author establishes a class of Trudinger-Moser inequality and proves the existence of the ground state solution to a class of Schrödinger equation with critical exponential growth. In addition, a class of quasilinear n-Laplace Schrödinger equations with degenerate potentials and of exponential growth is also studied. But to the best of our knowledge, the Schrödinger-Kirchhof equation that satisfies condition (V₁), (V₂) doesn’t seem to have been studied. Different from the first two results, the appearance of the term ∫ ℝ 2 | ∇ u | 2 d x , Some proof methods in the original text are invalid, so we have to find other methods, for the details see Lemmas 3.2 and 3.9.

Motivated by [17], we can prove the existence of the ground state solution to problem (1.5) as in [17]. In order to get the result we want, we use a version of Trudinger-Moser inequality.

Lemma 1.2. (Trudinger-Morse inequality [17] ) Assume that the potential V ( x ) ≥ 0 satisfies that V ( x ) = 0 at the ball B δ ( 0 ) centered at the origin with the radius δ and V ( x ) ≥ C 0 in ℝ 2 \ B 2 δ ( 0 ) for some δ > 0 . Then

sup u ∈ H 1 ( ℝ 2 ) , ∫ ℝ 2 | ∇ u | 2 + V ( x ) u 2 d x ≤ 1 ∫ ℝ 2 ( e 4 π u 2 − 1 ) d x < ∞ . (1.11)

Lemma 1.2 will be used to obtain the existence of ground state solution of the following Schrödinger-Kirchhof equation;

( − ( 1 + ∫ ℝ 2 | ∇ u | 2 d x ) Δ u + V ( x ) u = f ( u )     in   ℝ 2 , u ∈ H 1 ( ℝ 2 ) , (1.12)

Lemma 1.3. (Fatou’s Lemma) Let ( X , B , μ ) be a measure space, and { f n : X → [ 0, ∞ ] } be a sequence of non-negative measurable functions. Then the function lim inf n → ∞ f n is measurable and

∫ X lim inf f n d μ ≤ lim inf ∫ X   f n d μ . (1.13)

Now, we are ready to state the main results of this paper.

Theorem 1.4. Suppose that (V₁), (V₂) and f₀ - f₅ hold. If we further assume that

l i m t → + ∞ F ( t ) t 2 e α 0 t 2 = ∞ , (1.14)

then (1.12) admits a positive ground state solution.

2. Preliminaries

In this section, we give some useful notions and lemmas, which are used to prove our results.

Now, we introduce some notations. For any 1 ≤ r < ∞ , L r ( ℝ 2 ) is the usual Lebesgue space with the norm

‖ u ‖ r = ( ∫ ℝ 2 | u | r ) 1 r .

H 1 ( ℝ 2 ) is the usual Sobolev space with the norm

‖ u ‖ H 1 ( ℝ 2 ) 2 = ∫ ℝ 2 ( | ∇ u | 2 + | u | 2 ) d x .

Lemma 2.1. ( [17] ) Assume that u ∈ H 1 ( ℝ 2 ) such that

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x < + ∞ ,

where V ( x ) satisfies the assumption (V₁). Then there exists some constant c > 0 depending on δ and C 0 such that

∫ ℝ 2 | u | 2 d x < c ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x .

which was proved in [17].

Remark 2.2. If we define H V ( ℝ 2 ) as the completion of C 0 ∞ ( ℝ 2 ) under the norm

( ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x ) 1 2 ,

then Lemma 2.1 implies an result;

H V ( ℝ 2 ) = H 1 ( ℝ 2 ) .

The problem (1.5) associated functional is

I V ( u ) = 1 2 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + 1 4 ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 − ∫ ℝ 2   F ( u ) d x .

where F ( t ) = ∫ 0 t   f ( s ) d s , and its Nehari manifold is

N V ( u ) = { u ∈ H 1 ( ℝ 2 ) | u ≠ 0, 〈 I ′ V ( u ) , u 〉 = 0 } ,

where

〈 I ′ V ( u ) , u 〉 = ( 1 + ∫ ℝ 2 | ∇ u | 2 d x ) ∫ ℝ 2 | ∇ u | 2 d x + ∫ ℝ 2   V ( x ) | u | 2 d x − ∫ ℝ 2   f ( u ) u d x .

In order to study the problem (1.5) under the assumptions (V₁) and (V₂), we introduce the following limiting equation;

− ( a + b ∫ ℝ 2 | ∇ u | 2 d x ) Δ u + γ u = f ( u ) , (2.1)

where we recall from (V₂) that

sup x ∈ ℝ 2 V ( x ) = lim | x | → ∞ V ( x ) = γ > 0 ,

The corresponding functional and Nehari manifold associated with (2.1) are

I ∞ ( u ) = 1 2 ∫ ℝ 2 ( | ∇ u | 2 + γ | u | 2 ) d x + 1 4 ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 − ∫ ℝ 2   F ( u ) d x .

and

N ∞ = { u ∈ H 1 ( ℝ 2 ) | u ≠ 0 , 〈 I ′ ∞ ( u ) , u 〉 = 0 } ,

where

〈 I ′ ∞ ( u ) , u 〉 = ( 1 + ∫ ℝ 2 | ∇ u | 2 d x ) ∫ ℝ 2 | ∇ u | 2 d x + ∫ ℝ 2   γ | u | 2 d x − ∫ ℝ 2   f ( u ) u d x .

We can easily verify that if u ∈ N N , then

I V ( u ) = 1 2 ∫ ℝ 2 ( f ( u ) u − 2 F ( u ) ) d x − 1 4 ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 ,

and if u ∈ N ∞ , then

I ∞ ( u ) = 1 2 ∫ ℝ 2 ( f ( u ) u − 2 F ( u ) ) d x − 1 4 ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 .

3. The Proof of Theorem 1.3

In this section, we want to show that (1.5) has the existence of ground state solutions.

Lemma 3.1. N V and N ∞ are not empty.

Proof. we only prove that N V is not empty since the proof of N ∞ is similar. Let u 0 ∈ H 1 ( ℝ 2 ) be positive and compactly supported in a bounded domain Ω . Define

h ( t ) : = 〈 I ′ V ( t u 0 ) , t u 0 〉               = t 2 ∫ ℝ 2 ( | ∇ u 0 | 2 + V ( x ) | u 0 | 2 ) d x + t 4 ( ∫ ℝ 2 | ∇ u 0 | 2 d x ) 2                   − 2 ∫ ℝ 2   f ( t u 0 ) t u 0 d x ,   ∀ t > 0.

N V being not empty is a direct result of the fact; h ( t ) > 0 for t > 0 small enough and h ( t ) < 0 for t > 0 sufficiently large.

We first prove that h ( t ) > 0 for t > 0 small enough. From Remark 1.1, we conclude that for any ε > 0 , there exist C ε such that

| f ( t ) | ≤ ε | t | + C ε | t | μ ( e β 0 t 2 − 1 ) (3.1)

for any t ∈ ℝ . Using this estimate, we can write

h ( t ) = t 2 ∫ ℝ 2 ( | ∇ u 0 | 2 + V ( x ) | u 0 | 2 ) d x + t 4 ( ∫ ℝ 2 | ∇ u 0 | 2 d x ) 2 − 2 ∫ ℝ 2   f ( t u 0 ) t u 0 d x ≥ t 2 ∫ ℝ 2 ( | ∇ u 0 | 2 + V ( x ) | u 0 | 2 ) d x + t 4 ( ∫ ℝ 2 | ∇ u 0 | 2 d x ) 2     − 2 ε t 2 ∫ ℝ 2   u 0 2 d x − 2 | t | μ + 1 C ε ∫ ℝ 2   u 0 μ + 1 ( e β 0 ( t u 0 ) 2 − 1 ) d x , (3.2)

Which implies that h ( t ) > 0 for small t > 0 since μ > 4 .

Next, we prove that h ( t ) < 0 for t > 0 sufficiently large. The condition (f₂) implies that F ( t ) ≥ t μ F ( 1 ) when t > 1 . Then it follows that there exists C 3 > 0 such that F ( t ) ≥ t μ F ( 1 ) − C 3 for all t > 0 . this together with the condition (f₂) yields that there exists C , C 4 > 0 such that f ( t ) ≥ C t μ F ( 1 ) − C 4 for all t > 0 . Noticing that u 0 is compactly supported in the bounded domain Ω , we can write

h ( t ) = t 2 ∫ ℝ 2 ( | ∇ u 0 | 2 + V ( x ) | u 0 | 2 ) d x + t 4 ( ∫ ℝ 2 | ∇ u 0 | 2 ) 2 − 2 ∫ Ω   f ( t u 0 ) t u 0 d x ≤ t 2 ∫ ℝ 2 ( | ∇ u 0 | 2 + V ( x ) | u 0 | 2 ) d x + t 4 ( ∫ ℝ 2 | ∇ u 0 | 2 ) 2     − 2 F ( 1 ) C t μ + 1 ∫ Ω   u 0 μ + 1 d x + 2 C 4 ∫ ℝ 2   t u 0 d x , (3.3)

which implies that h ( t ) is negative for sufficiently large t > 0 .

Now, we set

m ∞ = inf { I ∞ ( u ) , u ∈ N ∞ }     and     m V = inf { I V ( u ) , u ∈ N V } ,

and we claim the following lemma.

Lemma 3.2. It holds that

0 < m V < m ∞ . (3.4)

Proof. To show that m V < m ∞ , it is enough to find u satisfying u ∈ N V such that I V ( u ) < m ∞ . From [18], we know that if

lim t → + ∞ F ( t ) t 2 e α 0 t 2 = ∞ ,

then m ∞ is attained by some w ∈ N ∞ . By the definition of V ( x ) , it is easy to check that

∫ ℝ 2 ( | ∇ w | 2 + V ( x ) | w | 2 ) d x + ( ∫ ℝ 2 | ∇ w | 2 d x ) 2 < ∫ ℝ 2 ( | ∇ w | 2 + γ | w | 2 ) d x + ( ∫ ℝ 2 | ∇ w | 2 d x ) 2 = ∫ ℝ 2   f ( w ) w d x .

Hence there exists t ∈ ( 0,1 ) such that

∫ ℝ 2 ( | ∇ ( t w ) | 2 + V ( x ) | t w | 2 ) d x + ( ∫ ℝ 2 | ∇ ( t w ) | 2 d x ) 2 = ∫ ℝ 2   f ( t w ) t w d x ,

which implies that u = t w ∈ N V . then it follows that

m V ≤ I V ( t w ) = 1 2 ∫ ℝ 2 ( | ∇ ( t w ) | 2 + V ( x ) | t w | 2 ) d x + 1 4 ( ∫ ℝ 2 | ∇ ( t w ) | 2 d x ) 2 − ∫ ℝ 2   F ( t w ) d x ≤ 1 2 ∫ ℝ 2 ( | ∇ ( t w ) | 2 + γ | t w | 2 ) d x + 1 4 ( ∫ ℝ 2 | ∇ ( t w ) | 2 d x ) 2 − ∫ ℝ 2   F ( t w ) d x = I ∞ ( t w ) ≤ max t ≥ 0 I ∞ ( t w ) = I ∞ ( w ) = m ∞

Next, we show m V > 0 . We prove this by contradiction. Assume that there exists some sequence u k ∈ N V such that I V ( u k ) → 0 , then we have

I V ( u k ) = I V ( u k ) − 1 4 〈 I ′ V ( u k ) , u k 〉 = 1 4 ‖ u k ‖ H V 2 + 1 4 ∫ ℝ 2 ( f ( u k ) u k − 4 F ( u k ) ) d x ≥ 1 4 ‖ u k ‖ H V 2 .

which implies that ‖ u k ‖ H V 2 → 0 . From u k ∈ N V and (3.1), we know that

∫ ℝ 2   f ( u k ) u k d x ≤ ∫ ℝ 2   ε | u k | 2 + C ε | u k | μ ( e β 0 u k 2 − 1 ) d x . (3.5)

By the Trudinger-Morse inequality (1.11) and μ > 4 , we get for any p > 1 ,

∫ ℝ 2 | u k | μ ( e β 0 u k 2 − 1 ) d x ≤ ( ∫ ℝ 2 | u k | μ p d x ) 1 p ( ∫ ℝ 2 ( e p ′ β 0 u k 2 − 1 ) d x ) 1 p ′ ≤ C ‖ u k ‖ H V μ . (3.6)

From (3.5) and (3.6), there exist C 1 , C 2 > 0 , such that

∫ ℝ 2   f ( u k ) u k d x ≤ C 1 ε ‖ u k ‖ H V 2 + C 2 ‖ u k ‖ H V μ .

Therefore

‖ u k ‖ H V 2 ≤ ∫ ℝ 2   f ( u k ) u k ≤ C 1 ε ‖ u k ‖ H V 2 + C 2 ‖ u k ‖ H V μ .

Since μ > 4 , there exist ρ > 0 such that

‖ u k ‖ H V 2 ≥ ρ > 0.

which is a contradicion to ‖ u k ‖ H V 2 → 0 .

We now consider a minimizing sequence { u k } k ⊂ N V for m V . Since

∫ ℝ 2 | ∇ ( | u k | ) | 2 d x ≤ ∫ ℝ 2 | ∇ ( u k ) | 2 d x ,

we can assume that u k ≥ 0 . The (A-R) condition (f₂), I V ( u k ) → m V > 0 and Remark 2.2 give that { u k } k is bounded in H 1 ( ℝ 2 ) , and then up to a subsequence, there exists u ∈ H 1 ( ℝ 2 ) , such that

u k ⇀ u   in   H 1 ( ℝ 2 ) ,   and   in   L p ( ℝ 2 )   for   any   p > 1 , u k → u   in   L l o c p ( ℝ 2 ) , u k → u   a .e .

By extracting a subsequence, if necessary, we define β , l ≥ 0 as

β = l i m k ∫ ℝ 2   f ( u k ) u k d x     and     l = ∫ ℝ 2   f ( u ) u d x .

By the weak convergence, it is obvious that l ∈ [ 0, β ] .

Lemma 3.3. It holds that β > 0 .

Proof. We proof this by contradiction. Assume that β = 0 . Then we have

I V ( u k ) = 1 2 ∫ ℝ 2 ( f ( u k ) u k − 2 F ( u k ) ) d x − 1 4 ( ∫ ℝ 2 | ∇ u k | 2 d x ) 2 d x → 0,

which contradicts (3.4).

Lemma 3.4 The case l = 0 cannot occur.

Proof. We prove this by contradiction. If l = 0 , then u = 0 , and u k → 0 in L l o c 2 ( ℝ 2 ) . we first claim that

lim k → + ∞ ∫ ℝ 2 ( γ − V ( x ) ) | u k | 2 d x = 0. (3.7)

For any fixed ε > 0 , we take R ε > 0 such that

| γ − V ( x ) | ≤ ε   for   any   | x | > R e .

combining this and the boundedness of u k in H 1 ( ℝ 2 ) , we derive that

∫ ℝ 2 ( γ − V ( x ) ) | u k | 2 d x = ∫ B R ε ( γ − V ( x ) ) | u k | 2 d x + ∫ B R ε c ( γ − V ( x ) ) | u k | 2 d x ≤ c ∫ B R ε | u k | 2 d x + K ε ,

where K = sup k ∫ ℝ 2 | u k | 2 d x . This together with u k → 0 in L l o c 2 ( ℝ 2 ) as k → + ∞ yields that

lim k → + ∞ ∫ ℝ 2 ( γ − V ( x ) ) | u k | 2 d x ≤ c ε ,

which implies that (3.7) hold.

Similarly to the proof of ( [19]. Proposition 6.1), we can get there exists some sequence t k ≥ 1 such that t k u k ∈ N ∞ and { t k } k converges to 1 as k → + ∞ .

Now, by (3.7), we can write

m ∞ ≤ lim k → + ∞ I ∞ ( t k u k ) = lim k → + ∞ ( I V ( t k u k ) + t k 2 2 ∫ ℝ 2 ( γ − V ( x ) ) | u k | 2 d x ) = lim k → + ∞ I V ( t k u k ) = lim k → + ∞ t k 2 ( 1 2 ∫ ℝ 2 ( | ∇ u k | 2 + V ( x ) | u k | 2 ) d x − ∫ ℝ 2 F ( t k u k ) t k 2 u k 2 u k 2 d x + t k 2 4 ( ∫ ℝ 2 | ∇ u k | 2 ) 2 ) .

This together with the monotonicity of F ( t ) t 2 and lim k → + ∞ t k = 1 gives

m ∞ ≤ lim k → + ∞ I V ( u k ) = m V ,

which contradicts (3.4). This accomplishes the proof of Lemma 3.4.

Note that (f₅) implies the following inequality;

1 − t 4 4 f ( u ) u + F ( t u ) − F ( u ) = ∫ t 1 [ f ( u ) u 3 − f ( s u ) ( s u ) 3 ] s 3 u 4 d s ≥ 0.   ∀ u ≠ 0,0 ≤ t ≤ 1. (3.8)

Lemma 3.5. If l = β , then u ∈ N V and I V ( u ) = m V .

Proof. If l = β , then

lim k → + ∞ ∫ ℝ 2   f ( u k ) u k d x → ∫ ℝ 2   f ( u ) u d x .

Then we can get

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 ≤ lim k → + ∞ ∫ ℝ 2 ( | ∇ u k | 2 + V ( x ) | u k | 2 ) d x + ( ∫ ℝ 2 | ∇ u k | 2 d x ) 2 = lim k → + ∞ ∫ ℝ 2   f ( u k ) u k d x = ∫ ℝ 2   f ( u ) u d x .

If the above equality holds, then u ∈ N V , and the lemma is proved. Therefore, it remains to show that the case where

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 < ∫ ℝ 2   f ( u ) u d x (3.9)

cannot occur. In fact, if (3.9) holds, we can take some t ∈ ( 0,1 ) such that t u ∈ N V . Indeed, let

g ( t ) = ∫ ℝ 2 ( | ∇ ( t u ) | 2 + V ( x ) | t u | 2 ) d x + ( ∫ ℝ 2 | ∇ ( t u ) | 2 d x ) 2 − ∫ ℝ 2   f ( t u ) t u d x ,

Obviously, g ( t ) is positive for small t. This together with g ( 1 ) < 0 implies that there exists t ∈ ( 0,1 ) such that g ( t ) = 0 , i.e., t u ∈ N V .

Using (f₅), we can obtain

f ( t u ) t u < t 4 f ( u ) u . (3.10)

From (3.8) we know that

F ( t u ) > t 4 − 1 4 f ( u ) u + F ( u ) . (3.11)

Combining (3.10) and (3.11), we derive

1 4 f ( t u ) t u − F ( t u ) < 1 4 f ( u ) u − F ( u ) . (3.12)

Since t ∈ ( 0,1 ) , by the define of N V ( u ) , (3.12) and Fatou’s lemma, we deduce that

m V ≤ I V ( t u ) − 1 4 〈 I ′ V ( t u ) , t u 〉 = t 2 4 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ∫ ℝ 2 ( 1 4 f ( t u ) t u − F ( t u ) ) d x < 1 4 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ∫ ℝ 2 ( 1 4 f ( u ) u − F ( u ) ) d x = I V ( u ) − 1 4 〈 I ′ V ( u ) , u 〉 ≤ lim k → ∞ [ I V ( u k ) − 1 4 〈 I ′ V ( u k ) , u k 〉 ] = m V .

Which is a contradiction.

In the following, we consider the case 0 < l < β . If 0 < l < β , then u k ⇀ u ≠ 0 in H 1 ( ℝ 2 ) . We can choose an increasing sequence { R j } j → + ∞ such that R j + 1 > R j + 1 ,

∫ B R j   f ( u ) u d x = l + o j ( 1 ) (3.13)

and

∫ B R j c | u | p d x = o j ( 1 )

for any 2 ≤ p < ∞ . We define

C j = B R j + 1 \ B R j = { x ∈ ℝ 2 | R j ≤ | x | < R j + 1 } .

Lemma 3.6. For the C j given above, we have

∫ C j   f ( u k ) u k d x = o j ( 1 ) (3.14)

and

∫ C j | ∇ u k | 2 d x = o j ( 1 ) (3.15)

Proof. We prove (3.14) by contradiction. If there exists some subsequence { j i } i of { j } such that (3.14) fails, then we must have

∑ i = 1 ∞   ∫ C j i   f ( u k ) u k d x = ∞ .

However, we have

∑ i = 1 ∞     ∫ C j i     f ( u k ) u k d x ≤ ∫ ℝ 2     f ( u k ) u k d x = ∫ ℝ 2 ( | ∇ u k | 2 + V ( x ) | u k | 2 ) d x + ( ∫ ℝ 2 | ∇ u k | 2 d x ) 2 < ∞ ,

which arrives at a contradiction. Similarly, we can also prove (3.15).

Lemma 3.7. ( [17] ) It holds that

lim k → + ∞ ∫ ℝ 2     F ( u k ) d x = ∫ ℝ 2     F ( u ) d x .

which was proved in [17].

Lemma 3.8. ( [20] ) Let Ω be a domain in ℝ N . Suppose { g n } , { h n } ⊂ L 1 ( Ω ) and h ∈ L 1 ( Ω ) . If

0 ≤ g n ≤ h n ,   g n ( x ) → 0,   h n → h     a .e .   x ∈ Ω

and

lim n → ∞ ∫ Ω     h n = ∫ Ω     h

then lim n → ∞ ∫ Ω     g n = 0 .

Lemma 3.9. It hold that

lim k → + ∞ ∫ B R j     f ( u k ) u k d x = ∫ B R j     f ( u ) u d x

provided j is large enough.

Proof. Since u k is bounded in H 1 ( ℝ 2 ) , there is C > 0 , and for any 2 ≤ p < + ∞ , we have

‖ u k ‖ L p ( B R j ) ≤ ‖ u k ‖ L p ( ℝ 2 ) ≤ ‖ u k ‖ H 1 ( ℝ 2 ) ≤ C . (3.16)

According to (ii) in definition 6.1 in reference [21], when m ( E ) < ∞ , it can be obtained

lim p → + ∞ ‖ f ‖ p = ‖ f ‖ ∞ . (3.17)

where E stands for measurable set.

From (3.16) and (3.17), we can deduce that

lim p → + ∞ ‖ f ‖ p = ‖ f ‖ ∞ ≤ C .

which implies that u k is bounded in B R j .

We can let g k = f ( u k ) u k − f ( u ) u , using (1.4), we can derive

| g k | = | f ( u k ) u k − f ( u ) u | ≤ | f ( u k ) u k | ≤ ε | u k | 2 + C ε | u k | μ + 1 ( e β 0 u k 2 − 1 ) .

Because u k is bounded in B R j , then there is M > 0 , we have

g k ≤ ε | u k | 2 + C ε | u k | μ + 1 ( e β 0 M − 1 ) = ε | u k | 2 + C 1 | u k | μ + 1 .

where C 1 = C ε ⋅ ( e β 0 M − 1 ) .

Using u k → u in L l o c p ( ℝ 2 ) and lemma 3.8, we can get

lim k → + ∞ ∫ B R j | f ( u k ) u k − f ( u ) u | = 0.

which implies that

lim k → + ∞ ∫ B R j     f ( u k ) u k d x = ∫ B R j     f ( u ) u d x . (3.18)

The proof is completed.

From (3.18) and Lemma 3.6, since R j + 1 < R j + 1 , we can extract a subsequence u k j such that for every j ∈ ℕ ,

∫ B R j     f ( u k j ) u k j d x = l + o j ( 1 )

and

∫ C j     f ( u k j ) u k j d x = o j ( 1 ) ,   ∫ C j | ∇ u k j | 2 d x = o j ( 1 ) ,   ∫ C j     u k j 2 d x = o j ( 1 ) .

Now, we take { u k j } as a new minimizing sequence renaming it { u j } j .

Lemma 3.10. It cannot be

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 < ∫ ℝ 2     f ( u ) u d x . (3.19)

Proof. If (3.19) is true, then there exists some t ∈ ( 0,1 ) such that t u ∈ N V . Since t ∈ ( 0,1 ) , by the define of N V ( u ) , (3.12) and Fatou’s lemma, we deduce that

m V ≤ I V ( t u ) − 1 4 〈 I ′ V ( t u ) , t u 〉 = t 2 4 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ∫ ℝ 2 1 4 f ( t u ) t u − F ( t u ) d x < 1 4 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ∫ ℝ 2 1 4 f ( u ) u − F ( u ) d x = I V ( u ) − 1 4 〈 I ′ V ( u ) , u 〉 ≤ lim k → ∞ [ I V ( u k ) − 1 4 〈 I ′ V ( u k ) , u k 〉 ] = m V .

which is a contradiction.

Lemma 3.11. It cannot be

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 > ∫ ℝ 2     f ( u ) u d x . (3.20)

Proof. We prove (3.20) by contradiction. If

∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 > ∫ ℝ 2     f ( u ) u d x . (3.21)

Since u j → u weakly in H 1 ( ℝ 2 ) , from Lemma 3.9, we can deduce

∫ ℝ 2   f ( u j ) u j d x + o j ( 1 ) = ∫ ℝ 2   f ( u ) u d x .

This implies that

‖ u ‖ H V 2 ≤ lim j → + ∞ ‖ u j ‖ H V 2 = lim j → + ∞ [ ∫ ℝ 2     f ( u j ) u j d x − ( ∫ ℝ 2 | ∇ u j | 2 d x ) 2 ] = ∫ ℝ 2     f ( u ) u d x − lim j → + ∞ ( ∫ ℝ 2 | ∇ u j | 2 d x ) 2 ≤ ∫ ℝ 2     f ( u ) u d x − ( ∫ ℝ 2 | ∇ u | 2 d x ) 2 < ‖ u ‖ H V 2 .

which is a contradiction.

End of the proof of Theorem 1.3. Lemma 3.10 and Lemma 3.11 imply I ′ V ( u ) = 0 . Hence

m V ≤ I V ( u ) − 1 4 〈 I ′ V ( u ) , u 〉 = 1 4 ∫ ℝ 2 ( | ∇ u | 2 + V ( x ) | u | 2 ) d x + ∫ ℝ 2 1 4 f ( u ) u − F ( u ) d x ≤ lim k → ∞ 1 4 ∫ ℝ 2 ( | ∇ u k | 2 + V ( x ) | u k | 2 ) d x + lim k → ∞ ∫ ℝ 2 1 4 f ( u k ) u k − F ( u k ) d x = lim k → ∞ [ I V ( u k ) − 1 4 〈 I ′ V ( u k ) , u k 〉 ] = m V .

which implies that u is a minimum point for I V on N V since u ≠ 0 . Therefore u is a ground state solution of the Equation (1.5) through the definition of the ground state.

4. Conclusion

In this paper, we use the Nehari manifold technique to prove the existence of ground state solutions for a class of Schrödinger-Kirchhoff equations with vanishing potential and exponential growth.

Acknowledgements

This work is supported by the Natural Science Foundation of China (11961081).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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